Problem: You, the user have 3 lives. In front of you are 5 cups - in each cup there is a piece of paper with 1 random number between 1 and 5 (inclusive) written on it.
You must guess the number in each cup, one at a time. If you get one wrong the numbers are randomized again and you must start from the beginning.
To win, you must guess all 5 numbers correctly before losing all 3 lives.
My question is: How do I calculate the odds of winning?
I think I've calculated that given there are 5 possible numbers for each of the 5 cups there are 53,130 different combinations.
Given that you have 3 lives, does that mean the odds of getting correct are 3 in 53,130?
I am not clear whether each cup has a different number or whether each cup has a $\frac{1}{5}$ chance of having any given number.
In the second case, the chance of correctly guessing the number in one cup is $\frac{1}{5}$, so the chance of correctly guessing the number in all five cups is $\frac{1}{5^5}=\frac{1}{3125}$. So the chance of losing a life in making an attempt is $1-\frac{1}{5^5}=\frac{3124}{3125}$. So the chance of losing three lives is $(1-\frac{1}{5^5})^3$. The chance of success is $1-(1-\frac{1}{5^5})^3=\frac{29287501}{30517578125}\approx0.1\%$.
In the first case, the chance of correctly guessing the first cup is $\frac{1}{5}$, the chance of correctly guessing the second is $\frac{1}{4}$ and so on. So the chance of correctly guessing them all is $\frac{1}{120}$. So the chance of losing a life is $\frac{119}{120}$. Hence the chance of losing three lives is $\left(\frac{119}{120}\right)^3=\frac{1685159}{1728000}$. So the chance of success is $1-\left(\frac{119}{120}\right)^3=\frac{42841}{1728000}\approx2.5\%$